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bezimeni [28]
3 years ago
12

You can increase the capacitance of a capacitor by A. Decreasing the plate spacing B. Increasing the plate spacing. ° C. Decreas

ing the area of the plates. D. Increasing the area of the plates. E. Both A and D F. Both B and C
Physics
1 answer:
eimsori [14]3 years ago
4 0

You can increase the capacitance of a capacitor by decreasing the plate spacing (A) or by increasing the area of the plates (D).

'A' and 'D' both do the job, so the correct choice is<em> (E)</em> .

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A square is 10 cm by 10 cm a student measure inside of the square is 9.9 cm and uses the measurements to calculate the area of t
Furkat [3]

Answer:1.5 2.15

Explanation:

2,5

5 0
3 years ago
A bowling ball with a mass of 7.0kg strikes a pin that had a mass of 2.0kg the pin flies forward with a velocity of 6.0m/s, and
Natalka [10]

The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.

We don't have external forces, so:

P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\

Where:

  • mb is the mass of the bowling ball
  • mp the mass of the pin
  • v_{0b}\quad and\quad v_{0p} the initial velocities of the bowling ball and the pin.
  • v_{1b}\quad and\quad v_{1p} the final velocities of the bowling ball and the pin.

Solving for v0b:

v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}

<h2>R/ The original velocity of the ball was 5.71 m/s.</h2>
6 0
3 years ago
A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to
White raven [17]

Answer:

1) \Delta s=1000\ ft

2)  \Delta s'=998.11\ ft.s^{-1}

3) t\approx125\ s

t'\approx463.733\ s

Explanation:

Given:

width of river, w=500\ ft

speed of stream with respect to the ground, v_s=8\ ft.s^{-1}

speed of the swimmer with respect to water, v=4\ ft.s^{-1}

<u>Now the resultant of the two velocities perpendicular to each other:</u>

v_r=\sqrt{v^2+v_s^2}

v_r=\sqrt{4^2+8^2}

v_r=8.9442\ ft.s^{-1}

<u>Now the angle of the resultant velocity form the vertical:</u>

\tan\beta=\frac{v_s}{v}

\tan\beta=\frac{8}{4}

\beta=63.43^{\circ}

  • Now the distance swam by the swimmer in this direction be d.

so,

d.\cos\beta=w

d\times \cos\ 63.43=500

d=1118.034\ ft

Now the distance swept downward:

\Delta s=\sqrt{d^2-w^2}

\Delta s=\sqrt{1118.034^2-500^2}

\Delta s=1000\ ft

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

v'=v.\cos37

v'=4\times \cos37

v'=3.1945\ ft.s^{-1}

<u>Now the net effective speed of stream sweeping the swimmer:</u>

v_n=v_s-v'

v_n=8-3.1945

v_n=4.8055\ ft.s^{-1}

<u>The  component of swimmer's velocity heading directly towards the opposite bank:</u>

v'_r=v.\sin37

v'_r=4\sin37

v'_r=2.4073\ ft.s^{-1}

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

\tan\phi=\frac{v_n}{v'_r}

\tan\phi=\frac{4.8055}{2.4073}

\phi=63.39^{\circ}

  • Now let the distance swam in this direction be d'.

d'\times \cos\phi=w

d'=\frac{500}{\cos63.39}

d'=1116.344\ ft

<u>Now the distance swept downstream:</u>

\Delta s'=\sqrt{d'^2-w^2}

\Delta s'=\sqrt{1116.344^2-500^2}

\Delta s'=998.11\ ft.s^{-1}

3)

Time taken in crossing the rive in case 1:

t=\frac{d}{v_r}

t=\frac{1118.034}{8.9442}

t\approx125\ s

Time taken in crossing the rive in case 2:

t'=\frac{d'}{v'_r}

t'=\frac{1116.344}{2.4073}

t'\approx463.733\ s

7 0
4 years ago
Pls pls help i am in 8th grade k12 How is the temperature affected (increased or decreased) when the iron block is placed in the
PtichkaEL [24]

Explanation:

the water temperature increase or decrease

6 0
3 years ago
Hardness may be defined as the resistance of a material to scratching.<br> a. True<br> b. False
OLga [1]

The correct answer its:

True

hope it helps

can I get the brainliest please?

5 0
4 years ago
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